题目原型：
Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).
For example:
Given binary tree {3,9,20,#,#,15,7},
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3
/ \
9 20
/ \
15 7
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return its bottom-up level order traversal as:
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[ [15,7]
[9,20],
[3],
] |
基本思路：
由于是从底向上，所以用到了栈。
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public ArrayList> levelOrderBottom(TreeNode root)
{ Stack> stack = new Stack>();
ArrayList> list = new ArrayList>();
ArrayList<TREENODE> nodeSet = new ArrayList<TREENODE>();
ArrayList<TREENODE> tmp ;
ArrayList<INTEGER> numSet ;
if(root!=null)
{
nodeSet.add(root);
while(nodeSet.size()>0)
{
tmp = new ArrayList<TREENODE>();
numSet = new ArrayList<INTEGER>();
//添加到stack中
for(TreeNode tn : nodeSet)
numSet.add(tn.val);
//添加到stack中
stack.push(numSet);
//求下一层的节点
for(TreeNode it : nodeSet)
{
if(it.left!=null)
tmp.add(it.left);
if(it.right!=null)
tmp.add(it.right);
}
nodeSet = tmp;
}
//添加到list中
while(stack.size()>0)
{
ArrayList<INTEGER> rs = stack.pop();
list.add(rs);
}
}
return list;
} |